Rotational Motion

Centre of Mass

For an object consists of $n$ point masses, we want to find a point mass that is equivalent to the object so that we do not have to do the Newton's second law for each point mass $$\sum^n_{i=1}m_i\vec{r}_i = M\vec{R}$$ So, the centre of mass $\textbf{R}$ is $$\bbox[5px,border:2px solid #666]{ \vec{R} = \frac{1}{M}\sum^n_{i=1} m_i\vec{r}_i }$$ When an external force is applied on one of the point masses $i$, the rate of change of its momentum is $$\dot{\vec{p}}_i = \vec{F}^{\text{(ext)}}_i + \sum^n_{j\neq i}\vec{F}_{ij},$$ where $\vec{F}^{\text{(ext)}}_i$ is the external force acting on the point mass and $\vec{F}_{ij}$ is the internal force due to the rest of the point masses. The sum of all the rate of change of momentum of the point masses is $$\sum^n_i\dot{\vec{p}}_i = \sum^n_i\vec{F}^{\text{(ext)}}_i + \sum^n_i\sum^n_{j\neq i}\vec{F}_{ij}.$$ Since $\vec{F}_{mn} = -\vec{F}_{nm}$ by Newton's third law, for some positive integers $m,n$, the second term cancelled. Hence, the total external force is $$\frac{d}{dt}\left(\sum^n_i\vec{p}_i\right)=\sum^n_im_i\ddot{\vec{r}}=\sum^n_i\vec{F}^{\text{(ext)}}_i$$ Substitute the definition of centre of mass, we have $$\sum^n_i\vec{F}^{\text{(ext)}}_i=M\ddot{\vec{R}}$$

Angular Displacement

An angle in radian is defined as $$\theta = \frac{s}{r}$$ where $s$ is the arc length and $r$ is the perpendicular distance between the point and rotational axis. The angular displacement is defined as $$\Delta \theta = \theta_2 - \theta_1$$ The angular velocity is defined as $$\omega = \lim_{\Delta \to 0}\frac{\Delta \theta}{\Delta t} = \frac{d\theta}{dt}$$ The angular acceleration is defined as $$\alpha = \lim_{\Delta \to 0}\frac{\Delta \omega}{\Delta t} = \frac{d\theta}{dt}$$

Rotational Kinetic Energy

Suppose a rigid body is rotating about a certain rotation axis. For a point of mass within the rigid body $$\begin{align} s &= r\theta \\ \frac{ds}{dt} &= \frac{d(r\theta)}{dt} \\ v &= r \frac{d\theta}{dt} \\ &= r\omega \end{align}$$ So the speed of a point in the rigid body is $$v = r\omega$$ Since the object is rigid, $\omega$ is the same for the whole object Then, the total kinetic energy of the rigid body is $$\begin{align} KE &= \sum^n_{i=1}\frac{1}{2} m_i v^2_i \\ &= \sum^n_{i=1} \frac{1}{2} m_i (r_i\omega)^2 \\ &= \frac{1}{2}\left(\sum^n_{i=1} m_i r_i^2\right) \omega^2 \\ \end{align}$$ The kinetic energy due to the rotational motion is then $$\bbox[5px,border:2px solid #666]{ KE_{rot} = \frac{1}{2}I \omega^2 }$$ where $$I = \left(\sum^n_{i=1} m_i r_i^2\right)$$ is called the moment of inertia. For continuous mass distribution, the moment of inertia is $$I = \int r^2 dm$$ As $r$ is measured from the rotation axis, moment of inertia is different with different rotational axis.

Example

A cube with uniform mass $M$ and side $a$ is put on a horizontal flat surface on one of its edge. It is then given a small push with negligible initial velocity and starts to fall. Assume the friction of the surface is sufficient such that the cube does not slide on the plane as it falls. Find the magnitude of the angular velocity of the cube at the instant when one face just strikes the surface.

Solution: Let the edge contacting the surface be the $x$-axis. The angular velocity vector $\vec{\omega}$ is $$\vec{\omega}=\omega\left( \begin{matrix} 1\\ 0\\ 0 \end{matrix} \right).$$

The moment of inerta about $\vec{\omega}$ is $$\begin{align} I &= \int r^2 dm = \int^a_0\int^a_0\int^a_0 (z^2+y^2)\left(\frac{M}{a^3}\right)dxdydz\\ &= \frac{M}{a^3}\int^a_0 \left[x\right]^{a}_{0}\left[\frac{z^3}{3}+y^2z\right]^{a}_{0}dy\\ &= \frac{M}{a^3}\int^a_0 \left(\frac{a^4}{3}+a^2y^2\right) dy\\ &= \frac{M}{a^3}\left[\frac{a^4y}{3}+\frac{a^2y^3}{3}\right]^{a}_{0}\\ &= \frac{M}{a^3}\frac{2a^5}{3}\\ &= \frac{2}{3}Ma^2. \end{align}$$ The rotational kinetic energy is $$T=\frac{1}{2}I\omega^2=\frac{1}{3}Ma^2\omega^2.$$ The potential energy loss is given by $$Mg\left(\frac{a}{\sqrt{2}}\right)(1-\cos\theta).$$ As there is no translational kinetic energy, by conservation of energy, $$Mg\left(\frac{a}{\sqrt{2}}\right)(1-\cos\theta)=\frac{1}{3}Ma^2\omega^2$$ When the cube face just strikes the surface, $\theta = \pi/4$. $$\frac{1}{2}Mga(\sqrt{2}-1)=\frac{1}{3}Ma^2\omega^2.$$ Therefore, $$\omega^2 = \frac{3}{2}\frac{g}{a}(\sqrt{2}-1).$$

Checkpoint

A cube with uniform mass $M$ and side $a$ is put on a horizontal flat surface on one of its edge. It is then given a small push with negligible initial velocity and starts to fall. Assume the surface is frictionless. Find the magnitude of the angular velocity of the cube at the instant when one face just strikes the surface.

Let the edge initially contacting the surface be the $x$-axis. The angular velocity vector $\vec{\omega}$ is $$\vec{\omega}=\omega\left( \begin{matrix} 1\\ 0\\ 0 \end{matrix} \right).$$ The moment of inerta about $\vec{\omega}$ is $$\begin{align} I &= \int r^2 dm = \int^{a/2}_{-a/2}\int^{a/2}_{-a/2}\int^{a/2}_{-a/2} (z^2+y^2)\left(\frac{M}{a^3}\right)dxdydz\\ &= \frac{M}{a^3}\int^{a/2}_{-a/2} \left[x\right]^{a/2}_{-a/2}\left[\frac{z^3}{3}+y^2z\right]^{a/2}_{-a/2}dy\\ &= \frac{M}{a^3}\int^{a/2}_{-a/2} \left(\frac{a^4}{12}+a^2y^2\right) dy\\ &= \frac{M}{a^3}\left[\frac{a^4y}{12}+\frac{a^2y^3}{3}\right]^{a/2}_{-a/2}\\ &= \frac{M}{a^3}\frac{2a^5}{12}\\ &= \frac{1}{6}Ma^2. \end{align}$$ The rotational kinetic energy is $$T=\frac{1}{2}I\omega^2=\frac{1}{12}Ma^2\omega^2.$$ As there is no horizontal force, the cube falls vertically $$\begin{align} z&=\frac{a}{\sqrt{2}}\cos\theta\\ \dot{z}&=\frac{a}{\sqrt{2}}\dot{\theta}\sin\theta\\ &=\frac{a}{\sqrt{2}}\omega\sin\theta \end{align}$$ Thus, the translational kinetic energy is $$\frac{1}{2}M\left(\frac{a}{\sqrt{2}}\omega\sin\theta\right)^2$$ The potential energy loss is given by $$Mg\left(\frac{a}{\sqrt{2}}\right)(1-\cos\theta).$$ By conservation of energy, $$Mg\left(\frac{a}{\sqrt{2}}\right)(1-\cos\theta)=\frac{1}{2}M\left(\frac{a}{\sqrt{2}}\omega\sin\theta\right)^2+\frac{1}{12}Ma^2\omega^2$$ $$\omega^2=\frac{\sqrt{2}g(1-\cos\theta)}{\frac{a}{2}\sin^2\theta+\frac{1}{6}a}$$ When the cube face just strikes the surface, $\theta = \pi/4$. Therefore, $$\begin{align} \omega^2 &= \frac{g}{a}\left(\sqrt{2}-\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\right)/\left(\frac{1}{2}\left(\frac{1}{\sqrt{2}}\right)^2+\frac{1}{6}\right)\\ &= \frac{12}{5}\frac{g}{a}(\sqrt{2}-1). \end{align}$$

Angular Momentum

Angular momentum is defined as $$\bbox[5px,border:2px solid #666] { \textbf{L} = \textbf{r} \times \textbf{p} }$$ The total angular momentum is $$\begin{align} \textbf{L} &= \sum^n_{i=1}\textbf{r}_i \times \textbf{p}_i \\ &= \sum^n_{i=1}m_i\textbf{r}_i \times \textbf{v}_i \\ \end{align}$$ If we consider only the magnitude of the angular momentum $$\begin{align} L &= \sum^n_{i=1}m_ir_{\bot i} v_i \\ &= \sum^n_{i=1}m_ir_{\bot i} (r_{\bot i} \omega) \\ &= \sum^n_{i=1}m_ir_{\bot i}^2 \omega \\ &= I\omega \end{align}$$ So, we conclude that the magnitude of an angular momentum is proportional to the magnitude of the angular speed $$\bbox[5px,border:2px solid #666] { L = I\omega }$$ Hence, rotational kinetic energy can also be expressed as $$KE_{rot}=\frac{1}{2}I\omega^2 = \frac{1}{2}\vec{\omega}\cdot\vec{L}$$

Inertia Tensor

From the definition of angular momentum $$\begin{align} \vec{L} &= \sum^n_{i=1}\vec{r_i} \times \vec{p_i} \\ &= \sum^n_{i=1}\vec{r_i} \times m_i \vec{v_i} \\ &= \sum^n_{i=1}m_i \vec{r_i} \times (\vec{\omega}_i \times \vec{r_i}) \\ &= \hat{x}\left[\omega_x\sum^n_{i=1}m_i(y^2_i + z^2_i) - \omega_y\sum^n_{i=1}m_ix_iy_i - \omega_z\sum^n_{i=1}m_ix_iz_i \right] \\ &+\hat{y}\left[-\omega_x\sum^n_{i=1}m_ix_iy_i + \omega_y\sum^n_{i=1}m_i(x^2_i + z^2_i) - \omega_z\sum^n_{i=1}m_iy_iz_i\right] \\ &+\hat{z}\left[-\omega_x\sum^n_{i=1}m_ix_iz_i - \omega_y\sum^n_{i=1}m_iy_iz_i + \omega_z\sum^n_{i=1}m_i(x^2_i + y^2_i)\right] \\ &= \left(\begin{matrix} \sum^n_{i=1}m_i(y^2_i + z^2_i) & - \sum^n_{i=1}m_ix_iy_i & - \sum^n_{i=1}m_ix_iz_i \\ -\sum^n_{i=1}m_ix_iy_i & -\sum^n_{i=1}m_i(x^2_i + z^2_i) & -\sum^n_{i=1}m_iy_iz_i \\ -\sum^n_{i=1}m_ix_iz_i & -\sum^n_{i=1}m_iy_iz_i & \sum^n_{i=1}m_i(x^2_i + y^2_i) \\ \end{matrix}\right) \left(\begin{matrix} \omega_x \\ \omega_y \\ \omega_z \\ \end{matrix}\right) \end{align}$$ The inertia tensor is defined as $$\textbf{I} = \left(\begin{matrix} \sum^n_{i=1}m_i(y^2_i + z^2_i) & - \sum^n_{i=1}m_ix_iy_i & - \sum^n_{i=1}m_ix_iz_i \\ -\sum^n_{i=1}m_ix_iy_i & -\sum^n_{i=1}m_i(x^2_i + z^2_i) & -\sum^n_{i=1}m_iy_iz_i \\ -\sum^n_{i=1}m_ix_iz_i & -\sum^n_{i=1}m_iy_iz_i & \sum^n_{i=1}m_i(x^2_i + y^2_i) \\ \end{matrix}\right)$$ So, we conclude that the angular momentum does not necessary lie with the same direction with the angular speed. However, they do lie in the same direction when the diagonal terms of the inertia tensor are all equal and the non-diagonal terms are all zero, or when the angular speed direction lies on one of the axis such that $\vec{\omega}$ has only one non-zero component (see example below).

The rotational kinetic energy can be then be express as $$KE_{rot}=\frac{1}{2}\vec{\omega}^{T}\textbf{I}\vec{\omega}$$

Example

A cube of uniform mass density with total mass $M$ and side $a$ is made to rotate about one of its edge. Find its angular momentum $\vec{L}$ and rotational kinetic energy.

Solution: Let the edge that the rotation axis passes through to be $x$-axis and set the origin at the middle of the edge. The angular velocity $\vec{\omega}$ is $$\vec{\omega} = \omega\left( \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right)$$ Let the inertia tensor to be $$\textbf{I}=\left(\begin{matrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \end{matrix}\right)$$ The angular momentum will be $$\vec{L} = \omega\left( \begin{matrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \end{matrix} \right) \left( \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\right) = \omega\left( \begin{matrix} I_{xx} \\ I_{yx} \\ I_{zx} \end{matrix}\right)$$ $$I_{xx}=\int^{a/2}_{-a/2}\int^0_{-a}\int^a_0(y^2+z^2)\rho dzdydx = \frac{2}{3}Ma^2$$ Since $x$ varies from $-a/2$ to $a/2$, $$I_{yx}=\int^{a/2}_{-a/2}\int^0_{-a}\int^a_0yx\rho dzdydx=0\\ I_{zx}=\int^{a/2}_{-a/2}\int^0_{-a}\int^a_0zx\rho dzdydx=0$$ The angular momentum is then $$\vec{L}=\omega\left( \begin{matrix} I_{xx} \\ I_{yx} \\ I_{zx} \end{matrix}\right)=\omega\left( \begin{matrix} \frac{2}{3}M a^2 \\ 0 \\ 0 \end{matrix}\right).$$ We see the angular momentum is just the moment of inertia multiplied by angular speed. The rotation kinetic energy is then $$\begin{align} T &= \frac{1}{2}\omega^T\vec{L}\\ &= \frac{1}{2} \left(\begin{matrix} \omega & 0 & 0 \end{matrix}\right) \left(\begin{matrix} \frac{2}{3}M a^2\omega \\ 0 \\ 0 \end{matrix}\right)\\ &= \frac{1}{3}Ma^2\omega^2 \end{align}$$

Checkpoint

A thin square plate of uniform mass density with total mass $M$ and side $a$ is made to rotate about an axis. Find its angular momentum $\vec{L}$ and rotational kinetic energy at this time.

Let $\rho$ be the mass density of the plate $$\rho = \frac{M}{a^2}.$$ As the rotation axis doesn't pass through the origin in this coordinate system, a new coodinate system which $y\to y-\frac{a}{2}$ is defined. The angular velocity vector $\omega$ can be written as $$\vec{\omega}=\frac{\omega}{\sqrt{(1/4)^2+(1/2)^2}}\left(\begin{matrix}1/4\\1/2\\0\end{matrix}\right)=\omega\left(\begin{matrix}1/\sqrt{5}\\2/\sqrt{5}\\0\end{matrix}\right)$$ So, $$\sum_im_i(y^2_i+z^2_i) = \int\int(y^2_i+z^2_i)dm = \int^a_0\int^{a/2}_{-a/2}(y^2+0^2)\rho dydx = \frac{1}{12}Ma^2\\ \sum_im_i(x^2_i+z^2_i) = \int\int(x^2_i+z^2_i)dm = \int^a_0\int^{a/2}_{-a/2}(x^2+0^2)\rho dydx = \frac{1}{3}Ma^2\\ \sum_im_i(x^2_i+y^2_i) = \int\int(x^2_i+y^2_i)dm = \int^a_0\int^{a/2}_{-a/2}(x^2+y^2)\rho dydx = \frac{5}{12}Ma^2\\ -\left(\sum_im_i(x^2_iy^2_i)\right) = -\left(\sum_im_i(y^2_ix^2_i)\right) = -\int^a_0\int^{a/2}_{-a/2}(xy)\rho dydx = 0 \\ $$ Since there is no $z$ volume, $$-\left(\sum_im_i(x^2_iz^2_i)\right) = -\left(\sum_im_i(z^2_ix^2_i)\right) \\= -\left(\sum_im_i(y^2_iz^2_i)\right) = -\left(\sum_im_i(z^2_iy^2_i)\right) = 0$$ So, the inertia tensor is $$\textbf{I} = \left( \begin{matrix} \frac{1}{12}Ma^2 & 0 & 0\\ 0 & \frac{1}{3}Ma^2 & 0\\ 0 & 0 & \frac{5}{12}Ma^2 \end{matrix} \right)$$ The angular momentum is then $$\vec{L}=\textbf{I}\vec{\omega}=\omega\left( \begin{matrix} \frac{1}{12}Ma^2 & 0 & 0\\ 0 & \frac{1}{3}Ma^2 & 0\\ 0 & 0 & \frac{5}{12}Ma^2 \end{matrix} \right)\left(\begin{matrix}1/\sqrt{5}\\2/\sqrt{5}\\0\end{matrix}\right)=\left( \begin{matrix} \frac{1}{12\sqrt{5}}M\omega a^2\\ \frac{2}{3\sqrt{5}}M\omega a^2\\ 0 \end{matrix} \right)$$ Hence, the rotational kinectic energy is $$\begin{align} T &= \frac{1}{2}\vec{\omega}^T\textbf{I}\vec{\omega}\\ &= \frac{1}{2}\left( \begin{matrix} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} & 0 \end{matrix} \right) \left( \begin{matrix} \frac{1}{12\sqrt{5}}Ma^2 \\ \frac{2}{3\sqrt{5}}Ma^2 \\ 0 \end{matrix} \right)\\ &=\frac{17}{120}Ma^2\omega^2 \end{align}$$

Torque

The derivative of angular momentum with respect to time is $$\begin{align} \frac{d\textbf{L}}{dt} &= \frac{d\textbf{r}}{dt} \times \textbf{p} + \textbf{r} \times \frac{d\textbf{p}}{dt}\\ &= \textbf{v} \times m\textbf{v} + \textbf{r} \times \textbf{F} \\ &= \textbf{r} \times \textbf{F} \end{align}$$ Torque is defined as $$\bbox[5px,border:2px solid #666] { \tau = \textbf{r} \times \textbf{F} }$$ Note: angular momentum is conserved as long as torque is zero, even the net force is not zero. For example, planet orbiting around a star has zero torque (since it is centripetal force, force is always parallel to displacement), so the angular momentum is conserved, but the net force is non-zero (gravitational force). $$\begin{align} \tau &= \textbf{r} \times \textbf{F}\\ &= \frac{d\textbf{L}}{dt} \\ &= \frac{d(I\vec{\omega})}{dt} \\ &= I\vec{\alpha} \end{align}$$ So, torque is also proportional to the rate of change of angular velocity $$\bbox[5px,border:2px solid #666] { \tau = I\vec{\alpha} }$$ For an object consists of $n$ point masses, the sum of the torque on all point masses is $$\sum^n_i\vec{\tau}_i = \sum^n_i\vec{r}_i\times\vec{F}^{\text{(ext)}}_i + \sum^n_i\sum^n_{j\neq i}\vec{r}_i\times\vec{F}_{ij},$$ where $\vec{F}_ij$ is the internal force between point mass $i$, $\vec{r}_i$ is the position vector from origin to the point mass $i$ and $j$ and $\vec{F}_i$ is the external force acting on point mass $i$. For two of the point masses, $m$ and $n$, by Newton's third law, $$\vec{F}_{mn}=-\vec{F}_{nm}$$ So, $$\vec{r}_m\times\vec{F}_{mn} + \vec{r}_n\times\vec{F}_{nm}=(\vec{r}_m-\vec{r}_n)\times\vec{F}_{mn}$$ If the internal force is parallel to the direction from point mass $m$ to $n$, then the term $$\sum^n_i\sum^n_{j\neq i}\vec{r}_i\times\vec{F}_{ij}=0.$$ Hence, $$\frac{d}{dt}\left(\sum^n_i\vec{L}_i\right) = \sum^n_i\vec{\tau}^{\text{(ext)}}_i.$$ When the total external torque is zero, $\sum^n_i\vec{L}_i$ is constant. This is not true when the internal force is not parallel to the direction between the point mass. One example can be magnetic force.

Rolling

If a uniform circular cylinder or sphere with radius $r$ rolls without slipping, since the arc changed is the same as distance of the centre of mass traveled $$\begin{align} x_{CM} &= s = r\theta\\ v_{CM} &= \frac{dx_{CM}}{dt} \\ &= \frac{d}{dt}(r\theta) \\ &= r\frac{d\theta}{dt} \\ &= r\omega \\ a_{CM} &= \frac{d}{dt}(r\omega) \\ &= r\frac{d\omega}{dt} \\ &= r\alpha \\ \end{align}$$

Example

A ring of radius $R$ and mass $M$ with uniform mass distribution is put above the ground with kinetic friction coefficient $\mu_k$. It is lifted slightly and given an angular speed $\omega_0$. It is then slowly put back to the ground so that the speed of its centre of mass is negligible. When will the ring stop slipping?

Solution: Define the direction of pointing out of the paper to be $z$-axis. The moment of inertia of the ring is $$I=\int r^2 dm = \int^{2\pi}_0 R^2\left(\frac{M}{2pi R}\right) Rd\phi=MR^2$$ The torque is $$\vec{\tau}=\mu_kMgR\hat{z} = I\dot{\vec{\omega}},$$ so $$\begin{align} \dot{\omega}&=-\frac{\mu_kg}{R}\\ \omega &= - \frac{\mu_kg}{R}t + \omega_0 \end{align}$$ The net force is $$\begin{align} \mu_kMg&=M\ddot{x}_{cm}\\ \dot{x}_{cm} &= \mu_kgt + C, \end{align}$$ for some constant $C$. At $t=0$, $\dot{x}_{cm}=0$, so $C = 0$ and $$\dot{x}_{cm}=\mu_kgt$$ When the wheel stop slipping, $\dot{x}_{cm}=R\omega$. So, $$\mu_kgt=R\left(-\frac{\mu_kg}{R}t+\omega_0\right).$$ Hence, the time is $$t=\frac{\omega_0R}{2\mu_kg}$$

Rotating Frame

Previously, we know that the acceleration of an object in polar plane coordinate system is given by $$\vec{a}_{inertial}(t) = (\ddot{r} - r\dot{\theta}^2)\hat{r} + (2r\dot{\theta} + r \ddot{\theta})\hat{\theta}$$ Suppose the reference frame is rotating at $\vec{\omega} = \omega \hat{z}$ $$\begin{align} \vec{a}_{inertial}(t) &= (\ddot{r} - r\dot{\theta}^2)\hat{r} + (2r\dot{\theta} + r \ddot{\theta})\hat{\theta} \\ &= \ddot{r}\hat{r} + r\omega^2 (-\hat{r}) + 2\dot{r}\omega\hat{\theta} + r\dot{\omega}\hat{\theta}\\ &= \ddot{r}\hat{r} + r\omega^2 (\hat{k}\times(\hat{k}\times\hat{r})) + 2\dot{r}\omega(\hat{k}\times\hat{r}) + r\dot{\omega}(\hat{k}\times\hat{r})\\ &= \ddot{r}\hat{r} + \omega\hat{k}\times(\omega\hat{k}\times r\hat{r}) + 2(\omega\hat{k}\times \dot{r}\hat{r}) + \dot{\omega}\hat{k}\times r\hat{r}\\ \end{align}$$ If the object does not have different angular velocity with the rotating reference frame, its velocity in the rotating reference frame is $$\vec{v} = \dot{r}\hat{r}$$ and its acceleration in the rotating reference frame is $$\vec{a} = \ddot{r}\hat{r}$$ So, $$\vec{a}_{inertial} = \vec{a} + \vec{\omega}\times(\vec{\omega}\times \vec{r}) + 2(\vec{\omega}\times \vec{v}) + \vec{\dot{\omega}}\times \vec{r}$$ Multiply by mass, we know that the force exerted on the object in the rotating reference frame is $$\bbox[5px,border:2px solid #666] { \vec{F} = \vec{F}_{real} - m\vec{\omega}\times(\vec{\omega}\times \vec{r}) - 2m(\vec{\omega}\times \vec{v}) - m\vec{\dot{\omega}}\times \vec{r} }$$ This is useful when the motion of the object is trivial in a rotating frame.

Example

A bug is moving with constant speed $v'$ on a turntable that is rotating with constant angular speed $\omega$. The bug moves along a circular path with radius $b$ and centre on the rotation axis of the turntable. The mass of the bug is $m$ and the static friction coefficient between the bug and the turntable is $\mu_s$. How fast can the bug moves before it starts to slip if it moves in the direction of the turntable rotation.

Solution: Use cylindrical coordinate with the rotation axis of the turntable as the $z$-axis. In the rotating turntable frame, the velocity of the bug is $v'\hat{\phi}$ and the force acted on the bug is $$\begin{align} \vec{F}'&=\vec{F}_{real}-2m\vec{\omega}\times\vec{v}'-m\dot{\vec{\omega}}\times\vec{r}'-m\vec{\omega}\times(\vec{\omega}\times\vec{r}')-m\vec{A}_0\\ -m\frac{v'^2}{b}\hat{r}' &= \vec{F}_{fric} + 2m\omega v'\hat{r} - (0) + m\omega^2b\hat{r}' - 0\\ \vec{F}_{fric} &= -m\left(2\omega v'+\omega^2b+\frac{v'^2}{b}\right)\hat{r}'\\ \mu_kmg &= m\left(2\omega v'+\omega^2b+\frac{v'^2}{b}\right)\\ v'&= \pm\sqrt{bg\mu_s}-\omega b \end{align}$$ As speed cannot be negative, the speed should be $$v'=\sqrt{bg\mu_s}-\omega b,$$ on the condition that $\omega\leq\sqrt{\frac{g\mu_s}{b}}$

Checkpoint

A bug is moving with constant speed $v'$ on a turntable that is rotating with constant angular speed $\omega$. The bug moves along a circular path with radius $b$ and centre on the rotation axis of the turntable. The mass of the bug is $m$ and the static friction coefficient between the bug and the turntable is $\mu_s$. How fast can the bug moves before it starts to slip if it moves in the direction opposite of the of the turntable rotation.

Use cylindrical coordinate with the rotation axis of the turntable as the $z$-axis. In the rotating turntable frame, the velocity of the bug is $-v'\hat{\phi}$ and the force acted on the bug is $$\begin{align} \vec{F}'&=\vec{F}_{real}-2m\vec{\omega}\times\vec{v}'-m\dot{\vec{\omega}}\times\vec{r}'-m\vec{\omega}\times(\vec{\omega}\times\vec{r}')-m\vec{A}_0\\ -m\frac{v'^2}{b}\hat{r}' &= \vec{F}_{fric} - 2m\omega v'\hat{r} - (0) + m\omega^2b\hat{r}' - 0\\ \vec{F}_{fric} &= -m\left(-2\omega v'+\omega^2b+\frac{v'^2}{b}\right)\hat{r}'\\ \mu_kmg &= m\left(-2\omega v'+\omega^2b+\frac{v'^2}{b}\right)\\ v'&= \pm\sqrt{bg\mu_s}+\omega b \end{align}$$ As speed cannot be negative, the speed should be $$v'=-\sqrt{bg\mu_s}+\omega b,$$ on the condition that $\omega>\sqrt{\frac{g\mu_s}{b}}$, or $$v'=\sqrt{bg\mu_s}+\omega b.$$